This 200-301 practice question tests your understanding of network infrastructure and connectivity. This is a configuration task: choose the command set that satisfies every stated requirement. Small differences — like 'secret' vs 'password' or 'transport input ssh' vs 'all' — change whether the answer is correct. A key principle to apply: a /26 subnet mask divides an IP address space into blocks of 64 addresses, calculated as 2^(32-26).. Once you have made your selection, read the full explanation to reinforce the concept and understand why each distractor is designed to mislead on exam day.
A host is configured with 192.168.60.33/26. Which address is the network address of its subnet?
192.168.60.0
This is correct because .33 is in the 0–63 /26 block.
192.168.60.32
Why wrong: This is wrong because /26 boundaries do not increment by 32.
192.168.60.64
Why wrong: This is wrong because .64 begins the next /26 block.
192.168.60.63
Why wrong: This is wrong because .63 is the broadcast address of the first /26 block, not the network address.
Answer choices
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
192.168.60.0
A /26 uses blocks of 64 addresses. In practical terms, the fourth-octet ranges are 0–63, 64–127, 128–191, and 192–255. Since 33 falls within the 0–63 block, the network address is 192.168.60.0. This is a straightforward boundary-identification question, but it catches people who memorize masks without understanding block sizes. The right approach is to find the correct block first, then take the first address in that block as the network address.
Key principle: A /26 subnet mask divides an IP address space into blocks of 64 addresses, calculated as 2^(32-26).
Answer analysis
For each option: why learners choose it and why it is or isn't the right answer here.
192.168.60.0
Why this is correct
This is correct because .33 is in the 0–63 /26 block.
Related concept
A /26 subnet mask divides an IP address space into blocks of 64 addresses, calculated as 2^(32-26).
192.168.60.32
Why it's wrong here
This is wrong because /26 boundaries do not increment by 32.
When this WOULD be correct
If the question asked for the first usable IP address in the subnet for 192.168.60.33/26, then option B (192.168.60.32) would be the correct answer, as it is the first address available for hosts in that subnet.
192.168.60.64
Why it's wrong here
This is wrong because .64 begins the next /26 block.
When this WOULD be correct
If the question asked for the network address of a subnet defined by a host IP of 192.168.60.65 with a subnet mask of /26, then option C (192.168.60.64) would be the correct answer, as it would represent the network address for that subnet.
192.168.60.63
Why it's wrong here
This is wrong because .63 is the broadcast address of the first /26 block, not the network address.
When this WOULD be correct
If the question asked for the broadcast address of the subnet containing the host 192.168.60.33/26, then 192.168.60.63 would be the correct answer, as it represents the highest address in that subnet range.
Option-by-option analysis
Understanding why wrong answers are wrong — and when they would be correct — is what separates a 750 score from a 900. The 200-301 exam frequently reuses these exact scenarios with slightly different constraints.
Why this is correct
This is correct because .33 is in the 0–63 /26 block.
Why this is wrong here
The /26 subnet mask has a block size of 64, not 32. The network addresses for /26 are 0, 64, 128, and 192. 192.168.60.32 is not a valid network address because it is not a multiple of 64.
★ When this WOULD be the correct answer
If the question asked for the first usable IP address in the subnet for 192.168.60.33/26, then option B (192.168.60.32) would be the correct answer, as it is the first address available for hosts in that subnet.
Why candidates choose this
Students often confuse /26 with /27, which has a block size of 32. Since 192.168.60.32 is a valid network address for a /27 subnet, they might mistakenly apply the same logic to a /26.
Why this is wrong here
192.168.60.64 is the network address of the next /26 subnet (64–127). The host address 192.168.60.33 belongs to the 0–63 range, not the 64–127 range.
★ When this WOULD be the correct answer
If the question asked for the network address of a subnet defined by a host IP of 192.168.60.65 with a subnet mask of /26, then option C (192.168.60.64) would be the correct answer, as it would represent the network address for that subnet.
Why candidates choose this
A test-taker might incorrectly calculate the subnet boundary by rounding up to the next multiple of 64, forgetting that the network address is the first address of the range containing the host.
Why this is wrong here
192.168.60.63 is the broadcast address of the 192.168.60.0/26 subnet, not the network address. The network address is always the first address in the subnet (all host bits set to 0).
★ When this WOULD be the correct answer
If the question asked for the broadcast address of the subnet containing the host 192.168.60.33/26, then 192.168.60.63 would be the correct answer, as it represents the highest address in that subnet range.
Why candidates choose this
Students sometimes confuse the broadcast address (last address) with the network address (first address), especially when the broadcast address ends in .63, which is close to the host address .33.
Analysis generated from the official 200-301blueprint and verified against question context. The “when correct” sections are what AI assistants cite when candidates ask “what’s the difference between these options?”
Common exam traps
Avoid confusing the first usable host address with the network address. Always identify the block range first.
Detailed technical explanation
Subnetting divides an IP network into smaller, manageable segments called subnets, each with its own network address. The subnet mask determines the size of each subnet by defining how many bits are used for the network portion versus the host portion. In this question, the host IP is 192.168.60.33 with a /26 mask, meaning 26 bits are fixed for the network and the remaining 6 bits are for hosts. This results in subnets with 64 IP addresses each (2^(32-26) = 64). To find the network address for a given host IP and subnet mask, you identify the subnet block that contains the host. With a /26 mask, the subnet blocks increment in steps of 64 in the last octet: 0–63, 64–127, 128–191, and 192–255. Since 33 falls within the 0–63 range, the network address is the first IP in that block, which is 192.168.60.0. This address identifies the subnet and is used by routers and switches to forward traffic correctly. A common exam trap is confusing the network address with other addresses in the subnet, such as the broadcast address or the first usable host address. For example, 192.168.60.32 is not a network boundary for /26 subnets; it is actually a host address within the 0–63 block. Understanding subnet block sizes and boundaries is critical to avoid such mistakes. Practically, network devices rely on the correct network address to route packets efficiently and prevent IP conflicts within subnets.
Key takeaway
A /26 subnet mask divides an IP address space into blocks of 64 addresses, calculated as 2^(32-26).
Real-world example
A network engineer segments a warehouse floor into three subnets: 20 scanners, 5 printers, and 2 management hosts. Picking the wrong mask wastes addresses or leaves too few usable hosts. Exam questions test whether you can apply CIDR notation, calculate block size, and identify the correct usable-host range for a given prefix.
What to study next
Review a /26 subnet mask divides an IP address space into blocks of 64 addresses, calculated as 2^(32-26)., then practise related 200-301 questions on the same topic to reinforce the concept.
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FAQ
Network Infrastructure and Connectivity — This question tests Network Infrastructure and Connectivity — A /26 subnet mask divides an IP address space into blocks of 64 addresses, calculated as 2^(32-26)..
The correct answer is: 192.168.60.0 — A /26 uses blocks of 64 addresses. In practical terms, the fourth-octet ranges are 0–63, 64–127, 128–191, and 192–255. Since 33 falls within the 0–63 block, the network address is 192.168.60.0. This is a straightforward boundary-identification question, but it catches people who memorize masks without understanding block sizes. The right approach is to find the correct block first, then take the first address in that block as the network address.
Review a /26 subnet mask divides an IP address space into blocks of 64 addresses, calculated as 2^(32-26)., then practise related 200-301 questions on the same topic to reinforce the concept.
A /26 subnet mask divides an IP address space into blocks of 64 addresses, calculated as 2^(32-26).
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Last reviewed: May 17, 2026
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